Explanation:
where is your diagram? lol
Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
Answer:
0.0605 Kg m^2
Explanation:
In this case where we have find he moment of inertia of this object about an axis perpendicular to the x-y plane and passing through the origin, we can just add three moment of inertia's .
MOI= 0.25×0.3^2 + 0.35×0.4^2- 0.45×0.2^2
= 0.0605 Kg m^2
Answer:
The Physical Behavior of Objects when Gravity is Missing
In order to be able to form a concept of the general physical conditions existing in a weightless state, the following must be noted: the force of the Earth's gravity pulling all masses down to the ground and thus ordering them according to a certain regularity is no longer active.
